【行間を読む】川上則雄・梁成吉「共形場理論と1次元量子系」p. 14(無限小座標変換による作用の変化)

キーワード

• 無限小座標変換

• エネルギー・運動量テンソル

該当箇所

場の理論の作用積分$${S}$$は無限小座標変換$${r_\mu\to r_\mu+\epsilon_\mu(\bm{r})}$$のもとで

$$
\delta S=-\int\frac{d^2r}{2π}T_{\mu\nu}(\bm{r})∂_\mu\epsilon_\nu(\bm{r})
$$

のように変化する。

解説

$${r^\mu\to r^\mu+\epsilon^\mu(\bm{r})}$$によって

$$
\begin{array}{rcl}\frac{\partial x'^\mu}{\partial x^\nu}&=&\left(\begin{matrix}1+\partial_1\epsilon^1 & \partial_1\epsilon^2\\\partial_2\epsilon^1 & 1+\partial_2\epsilon^2\end{matrix}\right),\\\frac{\partial x^\mu}{\partial x'^\nu}&=&\frac{1}{1+\partial_\mu\epsilon^\mu}\left(\begin{matrix}1+\partial_2\epsilon^2 & -\partial_1\epsilon^2\\-\partial_2\epsilon^1 & 1+\partial_1\epsilon^1\end{matrix}\right)+O(\epsilon^2),\\\frac{\partial}{\partial x'^\mu}&=&(1-\partial_\rho\epsilon^\rho)\left[\delta_{\mu1}\left((1+\partial_2\epsilon^2)\partial_1-\partial_1\epsilon^2\cdot\partial_2\right)\right.\\&&\qquad\qquad\left.+\delta_{\mu2}\left(-\partial_2\epsilon^1\cdot\partial_1+(1+\partial_1\epsilon^1)\partial_2\right)\right]+O(\epsilon^2)\end{array}
$$

となることに注意して、

$$
\begin{array}{rl}\delta S&=\int\frac{d^2r'}{2\pi}\mathcal{L}\left[A(\bm{r}'),\frac{\partial A(\bm{r}')}{\partial\bm{r}'},\bm{r}'\right]-\int\frac{d^2r}{2\pi}\mathcal{L}\left[A(\bm{r}),\frac{\partial A(\bm{r})}{\partial\bm{r}},\bm{r}\right]\\&=\int\frac{d^2r}{2\pi}\left( 1+\partial_\mu\epsilon^\mu \right)\mathcal{L}\left[A+\epsilon^\mu\partial_\mu A,(1-\epsilon^\rho\partial_\rho)\right.\\&\qquad\qquad\times\left(\partial_\mu+\delta_{\mu1}(\partial_2\epsilon^2\cdot\partial_1-\partial_1\epsilon^2\cdot\partial_2)+\delta_{\mu2}(\partial_1\epsilon^1\cdot\partial_2-\partial_2\epsilon^1\cdot\partial_1)\right)\\&\qquad\qquad\times\left.(A+\epsilon^\sigma\partial_\sigma A),\bm{r}+\bm{\epsilon}\right]-S+O(\epsilon^2)\\&=\int\frac{d^2r}{2\pi}\left[\mathcal{L}\partial_\mu\epsilon^\mu+\epsilon^\mu\partial_\mu A\cdot\frac{\delta \mathcal{L}}{\delta A}\right.\\&\qquad+\frac{\delta\mathcal{L}}{\delta\partial_\mu A}\left(-\partial_\rho\epsilon^\rho\cdot\partial_\mu A+\partial_\mu(\epsilon^\nu\partial_\nu A)\right.\\&\qquad\qquad\left.+\delta_{\mu1}(\partial_2\epsilon^2\cdot\partial_1A-\partial_1\epsilon^2\cdot\partial_2A)+\delta_{\mu2}(\partial_1\epsilon^1\cdot\partial_2A-\partial_2\epsilon^1\cdot\partial_1A)\right)\\&\qquad\left.+\epsilon^\mu\partial_\mu\mathcal{L}\right]+O(\epsilon^2)\\&=\int\frac{d^2r}{2\pi}\left[\partial_\mu(\epsilon^\mu\mathcal{L})+\epsilon^\mu\partial_\mu A\cdot\frac{\delta\mathcal{L}}{\delta A}+\partial_\mu\left( \epsilon^\nu\partial_\nu A\frac{\delta\mathcal{L}}{\delta\partial_\mu A} \right)-\epsilon^\mu\partial_\mu A\cdot\frac{\delta\mathcal{L}}{\delta A}\right.\\&\qquad\left.+\frac{\delta\mathcal{L}}{\delta\partial_\mu A}\left(-\delta_{\mu1}(\partial_1\epsilon^1\cdot\partial_1A+\partial_1\epsilon^2\cdot\partial_2 A)-\delta_{\mu2}(\partial_2\epsilon^2\cdot\partial_2A+\partial_2\epsilon^1\cdot\partial_1 A)\right)\right]\\&=\int\frac{d^2r}{2\pi}\left[\partial_\mu(\epsilon^\nu\delta^\mu_\nu\mathcal{L})-\partial_\mu\epsilon^\nu\cdot\frac{\delta\mathcal{L}}{\delta\partial_\mu A}\partial_\nu A\right]+\mathrm{surface}.\end{array}
$$

$${\mathcal{L}}$$が$${\bm{r}}$$に陽に依存しないとすると、

$$
\begin{array}{rl}\delta S&=\int\frac{d^2r}{2\pi}\partial_\mu\epsilon^\nu\left( \delta^\mu_\nu\mathcal{L}-\frac{\delta\mathcal{L}}{\delta\partial_\mu A}\partial_\nu A\right)+\mathrm{surface}\\&=-\int\frac{d^2r}{2\pi}\partial_\mu\epsilon^\nu T^\mu_\nu+\mathrm{surface}.\end{array}
$$

とできる。